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Let $(x_{i,j})$ be an infinite double sequence of nonnegative real numbers, and $ 0< p<1$.

I would like to know whether one can bound from above the sum \begin{equation} \sum_{i,j} x_{i,j}^p \end{equation} in terms of \begin{equation} \sum_{i}\Big(\sum_{j} x_{i,j}\Big)^p \quad , \quad \sum_{j}\Big(\sum_{i} x_{i,j}\Big)^p \quad \text{and} \quad \sum_{i,j} x_{i,j} \quad? \end{equation} The bound does not have to be tight, any upper bound will do.

The bound does not even have to be explicit, namely, proving the following statement would be useful. Let $\sum_{i,j} x_{i,j} = 1$, $\sum_{j} x_{i,j} = p_i$, and $\sum_{i} x_{i,j} = q_j$. If $\sum_{i} p_i^p < \infty$ and $\sum_{j}q_j^p < \infty$, does there exist a constant $M$ such that $\sum_{i,j} x_{i,j}^p \leq M$ for all sequences (bivariate probability distributions) $(x_{i,j})$ with marginals $(p_i)$ and $(q_j)$ ?

(This is a modified version of the problem posted a couple of days ago.)

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  • $\begingroup$ Is the upper bound allowed to use knowledge of how many $i$s and $j$s there are? $\endgroup$
    – Tom Leinster
    Commented Nov 3, 2012 at 23:51
  • $\begingroup$ Take $n\times n$ matrix with all entries equal to $a$. You want to bound $n^2a^p$ in terms of $n^{1+p}a^p$. It $a=n^{-1-\frac 1p}$, the second quantity is $1$ but the first is $n^{1-p}$. Voting to close. $\endgroup$
    – fedja
    Commented Nov 3, 2012 at 23:52
  • $\begingroup$ fedja: if $(x_{i,j})$ is an $n \times n$ matrix and the upper bound is allowed to mention $n$, then your comment doesn't show that the question is trivial. (On the evidence of your example, the first quantity could be $\leq$ $n^{1-p}$ times the max of the second and third.) It's true that this is being slightly generous to the original question, because "double sequence" suggests double infinite sequence, in which case your example does indeed answer it with a rather trivial "no". But it doesn't trivialize it for the finite case. $\endgroup$
    – Tom Leinster
    Commented Nov 4, 2012 at 1:23
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    $\begingroup$ Take $p_j = q_j = C^{-1} 1/j^\alpha$ with $1/p<\alpha<(2-p)/p$ and $C=\sum_j 1/j^\alpha$, so that $\sum_j q_j^p<\infty$. For any integer $n$, define $x_{i,j}=1/n^{\alpha+1}$ if $i,j \leq n$ and complete this family arbitrarily to have the correct marginals. Then $\sum_{i,j} x_{i,j}^p \geq n^{2 - p \alpha -p}$ which is not bounded as $n \to \infty$. $\endgroup$
    – Mikael de la Salle
    Commented Nov 8, 2012 at 8:46
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    $\begingroup$ Really, think a bit and stop just adding new restrictions. If I say that $\sum x_{ij}$ is completely irrelevant because it scales in a totally wrong way, it won't convince you, so let's do everything explicitly. Take an $n$ by $n$ matrix of $n^{-1-\frac 1p}$'s as before, which gives $\sum x_{ij}=n^{1-1/p}\to 0$. Add $x_{n+1,n+1}=1$. Then the first two sums are at $2$ exactly and the third tends to $1$. I hope you'll not ask me to prove that there may be no bound discontinuous at $(2,2,1)$... $\endgroup$
    – fedja
    Commented Nov 9, 2012 at 3:10

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